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# Using vectors,find the value of k such that the points $(k,-10,3),(1,-1,3)$ and $(3,5.3)$ are collinear.

$\begin{array}{1 1}(A)\;k=-1 \\ (B)\;k=-2 \\(C)\;k=2 \\ (D)\;k=0\end{array}$

Can you answer this question?

Toolbox:
• If two vector $\overrightarrow{a}$ and $\overrightarrow{b}$ are collinear, then $\overrightarrow{a}=\lambda \overrightarrow {b}$
Step 1:
Given $A(k, -10,3), B(1,-1,3), C(3,5,3)$
Let $\overrightarrow{OA}=(k\hat i-10\hat j+3\hat k)$, $\overrightarrow{OB}=(\hat i-\hat j+3\hat k)\;and\;\overrightarrow{OC}=(3\hat i+5\hat j+3\hat k)$
Let us first determine $\overrightarrow{AB}$
We know that $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$
$=(\hat i-\hat j+3\hat j)-(k\hat i-10\hat j+3\hat k)$
$=(1-k)\hat i+\hat j(10-1)$
$=(1-k)\hat i+9\hat j$
Let us determine $\overrightarrow{BC}$
$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$
$=(3\hat i+5\hat j+3\hat k)-(\hat i-\hat j+3\hat j)$
$=2 \hat i+6 \hat j$
Step 2:
Since the points A,B and C are collinear $\overrightarrow{AB}=\lambda \overrightarrow{BC}$
=>$\hat i(1-k)+9 \hat j=\lambda (2 \hat i+6 \hat j)$
(ie)$\hat i(1-k)+9 \hat j-\lambda (2 \hat i+6 \hat j)=0$
But $\hat i$ and $\hat j$ are non-colilinear hence
$(1-k-2 \lambda)=0$ and $9-6 \lambda=0$
Consider $9-6 \lambda=0$
$=> \lambda =\large\frac{9}{6}=\frac{3}{2}$
Substituting the value of $\lambda$ we get,
$1-k-2.\large\frac{3}{2}$$=0$
$-k-2=0$
$\qquad k=-2$
answered May 27, 2013 by